function inline with simple async code #59783
Description
function inline with simple async code
Future<void> foo(int i, int j) async{
await Future.delayed(const Duration(seconds: 1));
print(i);
print(j);
}
Future<void> main() async {
int i = 1;
int j = 1;
await foo(i, j);
}inline foo result:
Future<void> main() async {
int i = 1;
int j = 1;
await (int i, int j) async{
await Future.delayed(const Duration(seconds: 1));
print(i);
print(j);
}(i, j);
}expected code:
Future<void> main() async {
int i = 1;
int j = 1;
await Future.delayed(const Duration(seconds: 1));
print(i);
print(j);
}this pattern is valid but unnecessary it creates a lambda rather than inlining
await (int i, int j) async{ ................ }(i, j);